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[Edits in brackets]

Consider two finitely generated solvable minimax groups $G_i$ ($i = 1,2$) so that $1 \to N_i \to G_i \to Z_i \to 1$ [splits] with $N_i$ nilpotent, $Z_i$ infinite cyclic and $G_i$ are not themselves nilpotent. Assume $N_1$ and $N_2$ are isomorphic. Let $A_{k;i}$ be the quotients of the central series of $N_i$ and consider the matrices $M_{k;i}$ corresponding to the action of $Z_i$ on (the quotient of characteristic subgroups) $A_{k;i}$.

Question: which conditions on these matrices ensure that the groups $G_i$ are quasi-isometric?

e.g. is it sufficient [in the non-polycylic case] for them to have that their absolute Jordan form agree up to taking some powers? can the ratios of the powers [vary with] not be integer multiples of each other for different $k$? [how does one express known sufficient condition of the polycyclic cases in terms of these matrices?]

Remarks:

  • the classification of (such) groups up to quasi-isometry is (probably) still open. So I'm just looking for sufficient conditions on such groups which ensure they are quasi-isometric (not necessary conditions). Farb & Mosher have a classification of non-polycyclic finitely presented abelian-by-cyclic groups up to quasi-isometry, but I don't know if there is progress since then.

  • I don't know if the above set-up covers all solvable minimax groups [which are nilpotent-by-cyclic], but I'm looking at such groups first. [see comments]

  • by "the absolute Jordan form of $M_1$ and $M_2$ agree" I mean that there are $\alpha, \beta \in \mathbb{R}$" so that the [possibly complex] Jordan form of $M_2^\beta$ and $M_1^\alpha$ agree up to taking absolute values. (This condition is inspired from a condition whose origin [as a sufficient condition for the case $\mathbb{R}^n \ltimes \mathbb{R}$] I could not trace, but dates back at least to Farb & Mosher).

  • all solvable minimax groups are virtually-(locally nilpotent)-by-Abelian. So up to taking a finite index subgroup and [looking at a subgroup which has] only one element in the Abelian part, the above groups seem to be a not too ungeneric example.

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  • $\begingroup$ Comments on the remarks: (b) this construction covers nilpotent-by-cyclic solvable minimal groups, arbitrary solvable minimax groups are usually not virtually nilpotent-by-cyclic (even polycyclic ones). (d) actually all f.g. solvable minimax groups are virtually nilpotent-by-abelian. Essentially the restriction in this class of groups is that the nilpotent radical has codimension $\le 1$. This is a strong restriction (for example the polynomial Dehn function / exponential Dehn function / infinitely presented trichotomy is much simpler in this special case). $\endgroup$
    – YCor
    Commented Apr 21, 2021 at 12:47
  • $\begingroup$ Addressing the question "can the ratios of the powers...": In the case of SOL, there are non-commensurable lattices, and I think that it precisely comes from the fact that there are matrices in $\mathrm{SL}_2(\mathbf{Z})$ with trace $\ge 3$, which have no conjugate powers; $\endgroup$
    – YCor
    Commented Apr 21, 2021 at 12:49
  • $\begingroup$ Addressing the question in general, a necessary and sufficient condition might be "be cocompact lattices in the same group, that is a virtually solvable Lie group over a finite product of locally compact fields of characteristic zero". This is conjecturally true in the polycyclic case. $\endgroup$
    – YCor
    Commented Apr 21, 2021 at 12:58
  • $\begingroup$ @Ycor thanks for all the comments. For (b) I changed the question in the middle of writing and Z was originally Abelian (and not cyclic); this is why (b) is blatantly false now. To (d): thanks for the info! so if I replace Z by an Abelian group, this construction covers all solvable minimax groups? seems strange... (must be missing an important point). To "can the ratios..." I'm not too sure what you mean to say, seems rather to support the fact that the ratios of said powers must be multiple of each other...? $\endgroup$
    – ARG
    Commented Apr 21, 2021 at 15:30
  • $\begingroup$ It's not a "construction". But every f.g. virtually solvable minimax group is indeed virtually nilpotent-by-$\mathbf{Z}^d$ for some $d$ (for $d\ge 2$ it's not always virtually such a split extension). $\endgroup$
    – YCor
    Commented Apr 21, 2021 at 15:32

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