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This might be a silly question--but are there any examples of finitely generated subgroups of $\text{GL}_n(\mathbb{Q})$ that are known to not be virtually special?

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    $\begingroup$ Virtually special implies the much weaker PW property (acts properly on a CAT(0) cube complex). Plenty of f.g. Q-linear groups fail to have Property PW, e.g., those with Kazhdan's Property T (e.g. $\mathrm{SL}_3(\mathbf{Z})$), those with a distorted copy of $\mathbf{Z}$ such as $\mathrm{SL}_2(\mathbf{Z}[1/p])$ or (as already mentioned) solvable Baumslag-Solitar groups, non-abelien nilpotent torsion-free f.g. groups, etc. $\endgroup$
    – YCor
    Commented Jul 11 at 23:40
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    $\begingroup$ @HJRW no, it acts properly cocompactly on the product of a tree (Bruhat-Tits of $\mathrm{SL}_2(\mathbf{Q}_p)$ and the hyperbolic plane (symmetric space of $\mathrm{SL}_2(\mathbf{R})$). It doesn't have Property PW because it has a distorted infinite cyclic subgroup (inside a copy of a solvable Baumslag-Solitar group). $\endgroup$
    – YCor
    Commented Jul 13 at 22:26
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    $\begingroup$ @HJRW yes, sorry it's not cocompact (with a finite volume fundamental domain). $\endgroup$
    – YCor
    Commented Jul 14 at 7:02
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    $\begingroup$ @HJRW This is originally due to Haglund with a geometric argument, and I proved it by elementary means (in terms of commensurating actions, so eventually it's a fact about abstract actions of $\mathbf{Z}$ on sets) in my FW paper (Corollary 6.A.3). $\endgroup$
    – YCor
    Commented Jul 14 at 7:13
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    $\begingroup$ @HJRW, the proof by Haglund is Theorem 1.3 here: arxiv.org/pdf/0705.3386 $\endgroup$
    – Mark Hagen
    Commented Jul 18 at 12:12

1 Answer 1

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The Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a,b \mid bab^{-1}=a^n\rangle$ embeds into $\mathrm{GL}_2(\mathbb{Q})$, generated by the matrices $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}n&0\\0&1\end{pmatrix}$. For $n\ge 2$ it is not virtually special, thanks to translation length considerations. Thus, I think the "easiest" example of a group you're looking for is $\mathrm{BS}(1,2)$.

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    $\begingroup$ For clarity, perhaps I'll spell out the argument. (I got a bit confused above -- see comments -- so perhaps others may too.) Virtually special means "virtually embeds in a RAAG", without any cocompactness assumption. RAAGs act properly and cocompactly on their Salvetti complexes. In a group acting properly and cocompactly on a CAT(0) space, every infinite-order element has a positive translation-length, and in particular is undistorted. But every finite-index subgroup of $BS(1,n)$ has a distorted infinite-cyclic subgroup. $\endgroup$
    – HJRW
    Commented Jul 14 at 7:14

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