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What are sources of finitely generated $\mathbb C$-linear groups that are not $\mathbb Z$-linear?

Recall that a group is $R$-linear if it is isomorphic to a subgroup of $GL(n,R)$ for some $n$, where $R$ is a ring.

I know only one source: any solvable $\mathbb Z$-linear group is polycyclic. For example, the Baumslag-Solitar group $B(1,2)$ is solvable, $\mathbb C$-linear, and it contains dyadic rationals, and therefore, is not polycyclic (abelian subgroups of polycyclic groups are finitely generated).

My personal motivation for the question is an attempt to digest recent applications of virtual Haken conjecture implying that many $3$-manifold groups are $\mathbb Z$-linear.

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Igor: This is a very good question. Once can also ask for the further demarcation, namely between ${\mathbb Z}$-linear, ${\mathbb Q}$-linear and ${\mathbb R}$-linear f.g. groups. I thought aboyt this briefly in the context of solvable groups. Maybe the difference between real and rational cases is that the abelian quotient groups of the derived series are of finite rank in the rational case and infinite rank otherwise. –  Misha Feb 15 '13 at 0:07
One obvious obstruction is that f.g. groups which are ${\mathbb Q}$-linear have finite vcd, while ${\mathbb R}$-linear groups need not. On the other hand, all the ${\mathbb R}$-linear examples with infinite vcd I know, could be traced to solvable groups. –  Misha Feb 15 '13 at 0:50
Just to clarify: If you have a f.g. subgroup $G$ of $SL(n, {\mathbb Q})$, then you can have powers only of finitely many primes as denominators, thus, $G$ is in fact contained in some $S$-arithmetic group. The latter clearly has finite vcd. There is one more potential obstruction: Such groups $G$ have to be recursively presented (as subgroups of a f.p. group). On the other hand, I do not know of any examples of f.g. matrix groups without a recursive presentation. Maybe HW can help us here... –  Misha Feb 15 '13 at 1:34
Being a subgroup of a f.p. group is not an obstruction: if $G$ is a countable linear group over any field, then it embeds into a f.p. group with a solvable word problem, see Corollary 6.2 of "Infinitely Generated Subgroups of Finitely Presented Groups" by Baumslag-Cannonito-Miller. –  Igor Belegradek Feb 15 '13 at 4:52
@Misha and @Yves: Every finitely generated field is a finite extension of the field of rational functions $K(x_1,...,x_n)$ where $K$ is a the field of rationals (in the case of char=0) or the field with $p$ elements, $p$ prime. So every finitely generated linear group is inside a matrix group over the ring of rational functions $K(x_1,...,x_n)$. All that is covered in any first year graduate algebra course. Hence every finitely generated linear group is residually finite (Malcev),recursive and so on. –  Mark Sapir Feb 15 '13 at 13:44

2 Answers 2

You can use Margulis' super-rigidity theorem on $S$-arithmetic groups to say for example, that the group $SL_n({\mathbb Z}[\frac{1}{p})$ is never $\mathbb Z$-linear.

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For this group you have a infinitely generated abelian subgroup so superigidity is not needed, but this is a very good idea anyway: a torsion-free cocompact lattice in $SL_n(\mathbf{Q}_p)$ for $n\ge 3$ has finite cohomological dimension, its abelian subgroups are f.g. (this is a CAT(0) group) and superrigidity indeed shows it's not $\mathbf{Z}$-linear. –  YCor Feb 15 '13 at 8:11
NB it also works for cocompact irreducible lattices in $\mathrm{SL}_2(\mathbf{Q}_p)\times \mathrm{SL}_2(\mathbf{Q}_\ell)$, which are linear over $\mathbf{C}$ in dimension 2. –  YCor Feb 15 '13 at 9:20
True; but for this $SL_2({\mathbb Q}_p)\times SL_2({\mathbb Q}_l)$ group, the only lattices are co-compact, anyway. –  Venkataramana Feb 15 '13 at 11:37
Thank you, Aakumadula and Yves. –  Igor Belegradek Feb 15 '13 at 14:03

It is perhaps a stupid obstacle, but did you consider the torsion elements? For instance, you can have arbitrary order torsion elements in subgroups of $SL(2,\mathbb{R})$, while the finite orders of elements of $SL(2,\mathbb{Z})$ do not exceed 6.

By the way, do you want your group to be a discrete one, or it's a purely algebraic question?

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Victor, this is OK if you want subgroups of $SL(n,\mathbb{R})\smallsetminus SL(n,\mathbb{Z})$. The construction that the OP alludes to shows that discrete subgroups of $SL(2,\mathbb{C})$ are also contained in $SL(n,\mathbb{Z})$ for some large $n$. –  HJRW Feb 15 '13 at 7:58
Victor: the (usual) definition of linearity is just as I stated; there is no discreteness assumption and no restriction on $n$. So your obstruction seems not to answer the question. –  Igor Belegradek Feb 15 '13 at 12:07
@HW, @Igor: You are completely right, I'm sorry — I did not notice the "for some $n$" words. –  Victor Kleptsyn Feb 15 '13 at 13:02

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