The IA-congruence kernel of high rank free Metabelian groups

The congruence subgroup problem for a finitely generated group $Gamma$ and $Gleq Aut(Gamma)$ asks whether the map $hat{G}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(G,Gammaright)$? Here $hat{X}$ denotes the profinite completion of $X$. In this paper we investigate $Cleft(IA(Phi_{n}),Phi_{n}right)$, where $Phi_{n}$ is a free metabelian group on $ngeq4$ generators, and $IA(Phi_{n})=ker(Aut(Phi_{n})to GL_{n}(mathbb{Z}))$. We show that in this case $C(IA(Phi_{n}),Phi_{n})$ is abelian, but not trivial, and not even finitely generated. This behavior is very different from what happens for free metabelian group on $n=2,3$ generators, or for finitely generated nilpotent groups.