Consider two finitely generated solvable minimax groups $G_i$ ($i = 1,2$) so that $1 \to N_i \to G_i \to Z_i \to 1$ 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 ensure on these matrices ensure that the groups $G_i$ are quasi-isometric?

e.g. is it sufficient for them to have that their absolute Jordan form agree up to taking some powers? can the ratios of the powers not be integer multiples of each other for different $k$?

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 construction covers all solvable minimax groups, but I'm looking at such groups first.

• 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 only one element in the Abelian part, the above groups seem to be a not too ungeneric example.

[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.