The IA-congruence kernel of high rank free Metabelian groups

The congruence subgroup problem for a finitely generated group $\Gamma$ and $G\leq Aut(\Gamma)$ asks whether the map $\hat{G}\to Aut(\hat{\Gamma})$ is injective, or more generally, what is its kernel $C\left(G,\Gamma\right)$? Here $\hat{X}$ denotes the profinite completion of $X$. In this paper we investigate $C\left(IA(\Phi_{n}),\Phi_{n}\right)$, where $\Phi_{n}$ is a free metabelian group on $n\geq4$ 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.