The IA-congruence kernel of high rank free Metabelian groups

The congruence subgroup problem for a finitely generated group $\mathrm{\Gamma }$ and $G\le Aut(\mathrm{\Gamma })$ asks whether the map $\widehat{G}\to Aut(\widehat{\mathrm{\Gamma }})$ is injective, or more generally, what is its kernel $C(G,\mathrm{\Gamma })$? Here $\widehat{X}$ denotes the profinite completion of $X$. In this paper we investigate $C(IA({\mathrm{\Phi }}_n),{\mathrm{\Phi }}_n)$, where ${\mathrm{\Phi }}_n$ is a free metabelian group on $n\ge 4$ generators, and $IA(\Phi_{n})=\ker(Aut(\Phi_{n})\to GL_{n}($\mathbb{Z}$))$. We show that in this case $C(IA({\mathrm{\Phi }}_n),{\mathrm{\Phi }}_n)$ 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.