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corrected typos and corrected some important points (and tried not to make the comments obsolete in doing so)
ARG
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Quasi-isometry of solvable minimax groups

[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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