A metabelian group $G$ is determined up to isomorphism by "abelian data" $(A,M,\alpha)$ -- an abelian group $A:=G/[G,G]$, an $A$-module $M:=[G,G]$, and a cocycle $\alpha:A\times A\to M$ giving the extension $1\to M\to G\to A\to1$.
I wonder if one can define some $\alpha^{\wedge_{A,M}}$ fitting in $(M^\wedge,A^\wedge,\alpha^{\wedge_{A,M}})$, where $M^\wedge$ and $A^\wedge$ are the Pontryagin duals of $M$ and $A$ respectively, giving some group $G^\wedge$ with $1\to A^\wedge\to G^\wedge\to M^\wedge\to1$?
Particular cases which might be easier to handle: (1) when $G$ is class two nilpotent, the action of $A$ on $M$ is trivial, so $\alpha$ is just an abelian group cocycle; (2) when $\alpha$ is trivial, so that $G$ splits as a semidirect product $M\rtimes A$.
Note that if we drop the requirement $M=[G,G]$ and $\alpha$ is symmetric, i. e. $G$ is itself abelian, then we definitely do have such an $\alpha^{\wedge_{A,M}}$ since the above short exact sequence exists by "ordinary" Pontryagin duality.