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I'm wondering if anyone knows how to construct hyperbolic 3-manifolds whose fundamental group is RFRS. Clearly the recent work of Agol, Wise, etc. says that such manifolds are abundant, and in particular present in every commensurability class. But how do you construct examples?

The only examples of RFRS manifolds that I'm aware of are torus knot complements (thanks to Stefan Friedl for pointing this out), but these are of course non-hyperbolic.

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FYI, there's an error in my original paper. I stated that knot complements cannot be RFRS, but this is false e.g. for torus knots. A necessary condition is that Alexander polynomial has cyclotomic factors. Also, RAAGs are RFRS, and there are some graph 3-manifolds which are RAAGs when the defining graph is a tree (or $T^3$). – Ian Agol Oct 19 '12 at 4:00
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Agol's original paper on RFRS gives a nice short proof that the fundamental group of any manifold which also happens to be a finite-index subgroup of your favourite right-angled reflection group is RFRS. (Sketch proof: watch how loops bounce off the mirrors!)

So take your favourite right-angled reflection group $\Gamma$ in $\mathbb{H}^3$ and take the commutator subgroup $[\Gamma,\Gamma]$ (which is always torsion-free, as all the torsion injects into $H_1$). This will give you an example.

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Minor edit: iirc, Agol also assumes that the preimages of the mirrors are non-separating in the manifold, so you may need to pass to an additional cover to ensure that this holds too. – HJRW Oct 18 '12 at 14:41
Henry, thanks! So, in particular, are all the manifolds in your paper with Chesebro and DeBlois RFRS on the nose, without having to say "virtually"? – Dave Futer Oct 18 '12 at 15:04
Dave - not as far as I'm aware - indeed, we exhibit infinitely many commensurability classes of examples that are not commensurable with any 3-dimensional reflection group (see Section 6.3). So the above argument can't apply to those. – HJRW Oct 18 '12 at 20:55

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