Let $G$ be a "nice" infinite group: at least finitely presented and residually finite, maybe also linear and right-orderable (or even bi-orderable, or residually free nilpotent).

Consider an element $\lambda$ in the group ring $\mathbb Z[G]$ which is "residually invertible", ie every image $\overline\lambda\in\mathbb Z[G/H]$, $H$ a normal subgroup of finite index, is invertible. Is $\lambda$ itself invertible ?

Motivation: one can generalize the question to matrices in ${\rm M}_n(\mathbb Z[G])$, and also replace $\mathbb Z[G]$ by its Novikov completion in the direction of a nonzero morphism $u:G\to\mathbb R$ :

$$\mathbb Z[G]_u=\{\sum_{n=0}^\infty a_ng_n: a_n\in\mathbb Z,g_n\in G,u(g_n)\to+\infty\}.$$

A positive answer to the analogous question on detecting invertible matrices in ${\rm M}_n(\mathbb Z[G])_u$, for $G=\pi_1(M)$ with $M$ a closed $3$-manifold, would have the following consequence: a nonzero class $u\in H^1(M,\mathbb Z)={\rm Hom}(\pi_1(M),\mathbb Z)$ would be represented by a fibration $M\to S^1$ if and only if every twisted Alexander polynomial associated to a finite covering of $M$ is unitary (= bi-unitary).

Remarks. I have a proof for $\mathbb Z$ (!), elementary but not completely obvious. This implies the result (also for matrices) if $G$ is virtually free Abelian. From this one can prove the result for $G={\rm Heis}_3(\mathbb Z)$, the $3$-dimensional Heisenberg group over $\mathbb Z$ (matrices $\pmatrix{1&x&z\cr0&1&y\cr0&0&1}$ with $x,y,z\in\mathbb Z$), the simplest free nilpotent group. However, I do not see how to prove it for matrices over ${\rm Heis}_3(\mathbb Z)$, nor for general free nilpotent groups. Note that a proof for all free nilpotent groups would imply the result for residually free nilpotent groups, which include (I believe) most fundamental groups of closed $3$-manifolds, in particular all the hyperbolic ones.

  • $\begingroup$ Every hyperbolic 3-manifold has a finite-sheeted covering space whose fundamental group is residually torsion-free nilpotent. Is that the statement you're thinking of? $\endgroup$ – HJRW Oct 1 '14 at 21:06
  • $\begingroup$ Essentially yes. More precisely, it fibers over $S^1$ by a cebrated result of I. Agol, and a surface group is residually torsion-free nilpotent thus the whole group is also such. Sorry to have erased my upvote on your comment by a wrong move. $\endgroup$ – Jean-Claude Sikorav Oct 1 '14 at 21:41
  • $\begingroup$ Perhaps you should edit your question to indicate that it's not known to be the case that 'most 3-manifold groups' are residually free nilpotent group, then? $\endgroup$ – HJRW Oct 2 '14 at 5:58
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    $\begingroup$ surface groups are not residually free-nilpotent, so I wouldn't be optimistic about "most" fundamental groups of hyperbolic 3-folds. $\endgroup$ – YCor Oct 2 '14 at 18:15
  • $\begingroup$ I realize that I wrote "residually free nilpotent" but I meant "residually trosion free nilpotent". G. Baumslag proved in 1962 (On generalised free products) that an orientable surface group is residually torsion free nilpotent. $\endgroup$ – Jean-Claude Sikorav Dec 1 '16 at 18:53

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