The fundamental group $T(p,q)$ of the complement of a $(p,q)$torus knot (in $S^3$) admits the presentation $\langle a, b \mid a^p=b^q \rangle $. Is $T(p,q)$ linear, i.e., is there a faithful homomorphism of $T(p,q)$ into a matrix group over some field? For example, $T(2,3)$ and $T(3,3)$ are linear, since both can be embedded into the 3strand braid group $B_3$. (Indeed, $T(2,3)=B_3$.)
2 Answers
Torus knot groups are certainly linear. The universal cover of a torus knot complement is Euclidean space, and the action of the group on this cover factors through "translations on euclidean space" cross $PSL_2(\mathbb R)$. Both linear groups.
Generally it's conjectured that all 3manifold groups are linear. For all I know, there might be a proof already. See What is known about a 3manifold $M$ when its fundamental group is linear?

$\begingroup$ Many thanks. I already found now a paper that explains that in detail, namely: "Triangle groups, automorphic forms, and torus knots" by Valdemar V. Tsanov (see section 4.6. arxiv.org/abs/1011.0461 ). $\endgroup$ Oct 24, 2013 at 14:03

$\begingroup$ Hyperbolic groups were linear... at least as early as Heinz Hopf, I don't think Ian is taking credit for that. :) $\endgroup$ Oct 24, 2013 at 14:32

5$\begingroup$ Just to avoid confusion, Ryan's use of "hyperbolic groups" is for fundamental groups of manifolds of constant negative curvature, not for "Gromovhyperbolic groups". $\endgroup$– YCorOct 24, 2013 at 22:52
By the work of Agol, Liu, PrzytyckiWise and Wise we now know that fundamental groups of irreducible 3manifolds which are not closed graph manifolds are `virtually special', in particular they are linear over $\Bbb{Z}$.
Oddly enough, the only open case for linearity are fundamental groups of certain closed graph manifolds.

$\begingroup$ Its situations like this, and the earlier work of Waldhausen and Thurston on "sufficiently large" 3manifolds that give 3manifold theory a lovely uneven texture. It's a great study why one should not get too attached to only one tool. $\endgroup$ Oct 25, 2013 at 11:24