Is the fundamental group of any compact hyperbolic 3-manifold embeddable into a p-adic group?

Question

Is it true that for every compact hyperbolic $3$-manifold $M$ there exists a prime $p$, a finite field extension $K/\mathbb{Q}_p$, and an injective group homomorphism $$\tau \colon \pi_1(M) \to \mathrm{PSL}_2(\mathcal{O}_K) ?$$

Here $\mathcal{O}_K$ is the ring of integers of $K$.

This question is obviously related to the existence of faithful low dimensional representation of hyperbolic groups (see the following survey by Aschenbrenner, Friedl, and Wilton).

An affirmative solution will give a (partial) confirmation of Conjecture 1 from this paper by Luo.

A paper by Calegari and Emerton also discusses such group homomorphisms $\tau$.

Note that the answer is obviously 'yes' if $M$ is arithmetic.

Answer 1

I think this follows easily from a couple of facts. The discrete faithful representation of an orientable hyperbolic 3-manifold leads to an embedding $\pi_1(M) \to PSL_2(S) \subset PSL_2(k)$, where $k$ is a number field, and $S$ is a finitely-generated subring (generated e.g. by the coefficients of matrices of generators). For most prime ideals $\mathcal{P} \subset \mathcal{O}_k$, $S$ will have non-negative valuation, and hence will lie in the valuation ring $R_{\mathcal{P}} \subset k_{\mathcal{P}}$ (the ring of $\mathcal{P}$-adic integers), hence $\pi_1(M) \hookrightarrow PSL_2(R_\mathcal{P})$. Such a field is a finite extension of $\mathbb{Q}_p$, where $\mathcal{P}|p$. (I'm attempting to follow the notation and results of Chapter 0.7 of MacLachlan-Reid).

It's possible I've misinterpreted your notation. I'm interpreting your $K=k_\mathcal{P}, \mathcal{O}_K=R_\mathcal{P}$ (the valuation ring) to translate to MacLachlan-Reid's notation. If $\mathcal{O}_K$ does not denote the valuation ring, then I think you should clarify what you mean. This seems to me to be the only reasonable interpretation with $K=\mathbb{Q}_p, \mathcal{O}_K=\mathbb{Z}_p$.

Answer 2

This does not answer the question but at least gives a counterexample to the conjecture in Luo's paper.

I assume that $\mathrm{PSL}_2(R)$ is defined as the quotient of $\mathrm{SL}_2(R)$ by its center, the set of matrices $\begin{pmatrix}t & 0\\ 0 & t\end{pmatrix}$, $t^2=1$. This has the property $\mathrm{PSL}_2(\prod_i R_i)=\prod_i\mathrm{PSL}_2(R_i)$.

Lemma 1: let $G$ be a finitely generated group that is not embeddable into $\mathrm{PSL}_2$ over any field. Then $G\ast G$ is not embeddable into $\mathrm{PSL}_2$ over any commutative ring.

Proof: embed $G\ast G$ into $\mathrm{PSL}_2(R)$; we can suppose that $R$ is finitely generated; let $N$ be the nilradical of $R$, so $N^k=0$ for some $k$ and it follows that the kernel of $\mathrm{PSL}_2(R)\to\mathrm{PSL}_2(R/N)$ is a nilpotent group. Since $|G|\ge 3$, $G\ast G$ has no nontrivial nilpotent normal subgroup. Hence the composite map $G\ast G\to\mathrm{PSL}_2(R/N)$ is injective. Now $R/N$ is a reduced commutative noetherian ring and hence is embeddable in a finite product of fields (since $R/N$ has finitely many minimal prime ideals, whose intersection is trivial). So we have an embedding of $G\ast G$ into a product $\prod_{i=1}^n\mathrm{PSL}_2(K_i)$. Now we use a general property: if a group $H$ has the property that any two nontrivial normal subgroups have nontrivial intersection, then whenever $H$ embeds into a direct product of two groups, the projection to at least one factor remains injective. This applies to the group $G\ast G$. Hence it embeds into $\mathrm{PSL}_2(K_i)$ for some $i$, contradicting the assumption. $\Box$

Now I apply this to $G$ the binary icosahedral group of order 120. Indeed this group cannot be embedded in $\mathrm{PSL}_2$ of any field. This is easy, showing (in arbitrary characteristic) that, unlike in $G$, the centralizer of any element of order 2 in $\mathrm{PSL}_2(K)$ is solvable.

Since $G$ is the $\pi_1$ of a closed 3-manifold, so is $G\ast G$, namely of the connected sum of two copies of the Poincaré manifold.

(Btw, I expect that $G$ itself is not embeddable in $\mathrm{PSL}_2$ of any finite commutative ring. It true this is probably not too hard -- one can suppose that the ring is local and I think I've checked the case of odd characteristic.)

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Edit:

the binary icosahedral group $G$ of order 120 indeed cannot itself be embedded into $\mathrm{PGL}_2(R)$ for any commutative ring $R$.

Here by $\mathrm{PGL}_2(R)$ I mean the quotient of $\mathrm{GL}_2(R)$ by its subgroup of scalar matrices $\begin{pmatrix}\nu & 0\\ 0 & \nu\end{pmatrix}$, $\nu\in R^\times$.

First, since the only nontrivial proper normal subgroup of $G$ is its center $\{1,z\}$ of order 2, a homomorphism from $G$ to another group is injective iff $z$ is not in the kernel.

Suppose by contradiction that $G$ is embedded into $\mathrm{PGL}_2(R)$, $R$ a commutative ring. Let $T=\begin{pmatrix}a & b\\ c & a+d\end{pmatrix}$ be a lift of the image of $z$. That $z$ is not in the kernel means that $T$ is not a scalar matrix, that is $(b,c,d)\neq (0,0,0)$ in $R^3$. We can assume that $R$ is a finitely generated ring, and then since any finitely generated commutative ring is residually finite (more generally any f.g. module over a noetherian commutative ring is residually of finite length), we can suppose that $R$ is finite. Since $R$ is then a product of local rings, we can suppose that $R$ is local, with maximal ideal $M$ and residual field $K=R/M$.

Since by the previous part of the post, $G$ cannot be embedded into $\mathrm{PGL}_2$ of a field (the argument being that in $\mathrm{PGL}_2(K)$, the centralizer of any nontrivial element is solvable), $T$ has to belong to the kernel of the map to $\mathrm{PGL}_2(K)$, so $(b,c,d)\in M^3$. Then we can replace $R$ by a proper quotient so as to assume that $Mb=Mc=Md=0$, and that $Rb+Rc+Rd$ has dimension 1 as vector space over $R/M$. That is, we can write $(b,c,d)=(\beta r,\gamma r,\delta r)$ with $r\in R$ with annihilator $M$ and $(\beta,\gamma,\delta)\in K^3\smallsetminus\{(0,0,0)\}$. To get a contradiction, let us check that the centralizer of $T$ in $\mathrm{PGL}_2(R)$ is solvable. It is enough to show that its inverse image $H$ in $\mathrm{GL}_2(R)$ is solvable. We have $H=\{h\in\mathrm{GL}_2(R):\exists q\in R^\times: hTh^{-1}=qT\}$. Note that since $a\in R^\times$, $q$ is unique, and the resulting map $H\to R^\times$, $h\mapsto q$ is a group homomorphism, whose kernel is the centralizer of $T$ in $\mathrm{GL}_2(R)$. But this centralizer is also the inverse image in in $\mathrm{GL}_2(R)$ of the centralizer of the matrix $\begin{pmatrix}0 & \beta\\ \gamma & \delta\end{pmatrix}$ in $\mathrm{GL}_2(K)$, and this is solvable since the latter matrix is not a scalar matrix.

Since $G$ is not solvable and $z$ is central, this is a contradiction.