# Solvable groups that are linear over $\mathbb{C}$ but not over $\mathbb{Q}$?

Let $$\Gamma$$ be a finitely generated finitely presented virtually solvable group. Assume that there exists an injective representation $$\Gamma \to \operatorname{GL}_n(\mathbb{C})$$. Is it true that there always exists an injective representation $$\Gamma \to \operatorname{GL}_m(\mathbb{Q})$$ for some $$m$$?

A natural way of arguing that comes to my head is the following. Let $$R_n(\Gamma)$$ be the representation scheme of $$\Gamma$$ in $$\operatorname{GL}_n$$ (that is, the subscheme in $$\operatorname{GL}_n^{\times s}$$, where $$s$$ is the number of generators, which is defined by the relations viewed as algebraic equations on matrix elements). Inside it, we have the subset of injective representations $$R^\text{inj}_n(\Gamma) \subseteq R_n(\Gamma)$$. If it were an algebraic subvariety, one could say that since $$R^\text{inj}_n(\Gamma)$$ contains a point over $$\mathbb{C}$$, it contains a point over a number field $$K$$. Thus $$\Gamma$$ has an injective representation with values in $$\operatorname{GL}_n(K)$$ which by restriction of scalars gives rise to an injective representation with values in $$\operatorname{GL}_m(\mathbb{Q})$$ with $$m=n[K:\mathbb{Q}]$$.

The problem here is that $$R^\text{inj}_n(\Gamma)$$ is not a Zariski open subset. One has to take out the subvarieties that correspond to all possible kernels of representations of rank $$n$$ and this is a countable union of subvarieties. Even in the simplest case when $$\Gamma=\mathbb{Z}$$ and $$n=1$$ one should take out all the roots of unity (that is, $$R^\text{inj}_n(\Gamma)=\mathbb{G}_m\setminus \mu_{\infty}$$).

Notation. For $$G$$ a group and $$x$$, $$y$$ two elements of $$G$$, we set $$x^y := y^{-1}x y$$ and $$[x, y] := x^{-1}y^{-1}xy$$.

Definition. A group is said to have finite Prüfer rank $$r$$ if every finitely generated subgroup can be generated by $$r$$ elements and $$r$$ is the least such integer.

The answer to the question stated in the body of the post is no, as witnessed by the following finitely presented counter-example due to Baumslag and Remeslennikov [1, Theorem 11.1.5]:

Claim. Let $$G = \langle a, s, t \, \vert \, a^t = a a^s, [a, a^s] = 1 = [s,t] \rangle$$ be the Baumslag–Remeslennikov group. Then $$G$$ is a finitely presented torsion-free metabelian group with subgroup $$\langle a, s \rangle \simeq \mathbb{Z} \wr \mathbb{Z}$$.

Sketch of the claim's proof. Check that $$[G, G] = \langle a^{s^i} \,\vert\, i \in \mathbb{Z} \rangle \simeq \mathbb{Z}^{(\mathbb{Z})}$$. It follows immediately that $$G$$ is torsion-free and metabelian. See [1, Proof of Theorem 11.1.5] for details.

Note. More generally, a theorem of Baumslag and Remeslennikov (1973) states that every finitely generated metabelian group can be embedded into a finitely presented metabelian group. Even more to the point is Thomson's theorem (1977) [2]: every finitely generated linear soluble group can be embedded into a finitely presented linear soluble group. Starting from $$H = \mathbb{Z} \wr \mathbb{Z}$$, Thomson's theorem yields a linear and finitely presented soluble group $$G$$ containing $$H$$. This group is made explicit in the above claim.

Corollary. The Baumslag–Remeslennikov group, i.e., the group $$G$$ of the above claim, is $$\mathbb{C}$$-linear but not $$\mathbb{Q}$$-linear.

The proof of the previous corollary relies on the following two lemmas.

Lemma 1. (YCor's key argument) A finitely generated soluble $$\mathbb{Q}$$-linear group has finite Prüfer rank.

Lemma 2. (Levic–Remeslennikov, 1969 [3] and [4]) A finitely generated torsion-free metabelian group is $$\mathbb{C}$$-linear.

Proof of Lemma 1. Because of Mal'cev's Theorem (1951), see e.g. [1, Theorem 3.1.6.ii], it suffices to show that any finitely generated subgroup of $$T_n(\overline{\mathbb{Q}})$$ ($$n \ge 1$$), the group of invertible upper $$n \times n$$ triangular matrices over the algebraic closure of $$\mathbb{Q}$$, has finite Prüfer rank. Such a subgroup $$H$$ is a finitely generated subgroup of $$T_n(K)$$ for some number field $$K = K(H)$$. It is therefore the extension of a subgroup $$N$$ of $$U_n(K)$$, the $$n \times n$$ unipotent matrices over $$K$$ by a finitely generated subgroup $$Q$$ of $$(K^{\times})^n = (K \setminus \{0\})^n$$. Since $$Q$$ is a finitely generated Abelian group, it has finite Prüfer rank and so has $$N$$ since the additive group of $$K$$ is a finite dimensional $$\mathbb{Q}$$-linear space. Now, it is an easy exercise to show that the property of having finite Prüfer rank is stable under taking extensions.

Proof of the corollary. The Baumslag–Remeslennikov group is $$\mathbb{C}$$-linear by the Claim and Lemma 2. It cannot be $$\mathbb{Q}$$-linear because of Lemma 1. Indeed, it contains a subgroup of infinite Prüfer rank, namely $$\langle a, s \rangle \simeq \mathbb{Z} \wr \mathbb{Z} \supset \mathbb{Z}^{(\mathbb{Z})}$$.

Note. As pointed out by YCor in a comment, a simple and direct way to show that the Baumslag-Remeslennikov group $$G$$ is linear is to leverage the semi-direct decomposition $$G \simeq \mathbb{Z}^{(\mathbb{Z})} \rtimes \mathbb{Z}^2 \subset \mathbb{Z}[s^{\pm 1}] \rtimes (\mathbb{Z}[s^{\pm 1}, t^{\pm 1}])^{\times}.$$ The Magnus Embedding Theorem (1939), see e.g. [1, 11.3.2] may serve a similar purpose in more general situations.

• [1] J. Lennox and D. Robinson, "The Theory of Infinite Soluble Groups", 2004.
• [2] M. Thomson, "Subgroups of Finitely Presented Solvable Linear Groups", 1977.
• [3] E. Levic, "Representation of soluble groups by matrices over a certain field of characteristic zero", 1969.
• [4] V. Remeslennikov, "Representation of finitely generated metabelian groups by matrices", 1969.
• For Lemma 2: in the particular case of a group of the form $G\ltimes R$ with $R$ a domain and $G\subset R^*$, one has the obvious linear representation of the form $\begin{pmatrix} g& r \\ 0 & 1\end{pmatrix}$, $g\in G$, $r\in R$. This is even more elementary than Magnus' embedding.
– YCor
Nov 17, 2023 at 6:16
• @LSpice Very right, thanks ! Fixed. Nov 17, 2023 at 6:39

$$\DeclareMathOperator\GL{GL}$$No, it's not true that every finitely generated solvable group that is linear over $$\mathbf C$$ is also linear over $$\mathbf Q$$. For instance, f.g. solvable subgroups of $$\GL_n(\mathbf{Q})$$ have finite Prüfer rank, but not all f.g. solvable subgroups of $$\GL_n(\mathbf{C})$$.

For instance, the wreath product $$\mathbf{Z}\wr\mathbf{Z}$$ is not $$\mathbf{Q}$$-linear, although it is linear in dimension $$2$$: fixing any transcendental element $$t$$, it can be identified to the subgroup $$\left\{\begin{pmatrix}t^n &P(t)\\0&1\end{pmatrix}:n\in\mathbf{Z},\;P\in\mathbf{Z}[t,t^{-1}]\right\}.$$

To make the whole self-contained: if $$G$$ is a solvable group and $$[G,G]$$ contains copies of $$\mathbf{Z}^d$$ for unbounded $$d$$ then $$G$$ is not linear over $$\mathbf{Q}$$.

Proof: by contraposition: suppose that $$G$$ is linear over $$\mathbf{Q}$$, say $$G\subset\GL_n(\mathbf{Q})$$. Passing to a finite index subgroup, $$G$$ has a connected Zariski closure. Hence there is a finite extension $$L$$ of $$\mathbf{Q}$$ such that $$G$$ is conjugate into the upper triangular subgroup of $$\GL_n(L)$$. So $$[G,G]$$ is isomorphic to a subgroup of the upper unipotent (= upper triangular with identity diagonal) subgroup of $$\GL_n(L)$$. The latter is an iterated extension of $$n(n-1)/2$$ copies of $$L$$. Hence its free abelian subgroups have bounded rank, say $$\le \dim_{\mathbf{Q}}(L)n(n-1)/2$$.

(For the above matrix group $$\mathbf{Z}\wr\mathbf{Z}$$, the derived subgroup is the ideal $$\{P\in\mathbf{Z}[t,t^{-1}]:P(1)=0\}$$.)

• More generally, given a field $K$ and $n\ge 1$, the wreath product $\mathbf{Z}^n\wr \mathbf{Z}=\mathbf{Z}[t_1^{\pm 1},\dots,t_n^{\pm 1}]\rtimes \langle t_1,\dots,t_n\rangle$ is linear over $K$ if and only $K$ has characteristic zero and transcendence degree $\ge n$ over $\mathbf{Q}$.
– YCor
Nov 16, 2023 at 11:47
• What if one would ask instead, whether each $\mathbb{C}$-linear group is residually $\mathbb{Q}$-linear? Nov 16, 2023 at 13:05
• @HenrikRüping a f.g. $\mathbf{C}$-linear group is residually finite, hence residually $\mathbf{Q}$-linear (f.g. can't be dropped for free since $\mathrm{PSL}_2(\bar{\mathbf{Q}})$ is not residually $\mathbf{Q}$-linear). Your question might become more interesting if you ask "residually linear in bounded dimension".
– YCor
Nov 16, 2023 at 13:35