The Congruence Subgroup Problem for the Free Metabelian group on $ngeq4$ generators

The congruence subgroup problem for a finitely generated group $Gamma$ asks whether the map $hat{Autleft(Gammaright)}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(Gammaright)$? Here $hat{X}$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $Cleft(mathbb{Z}^{n}right)=left{ 1right}$ for every $ngeq3$, but $Cleft(mathbb{Z}^{2}right)=hat{F}_{omega}$, where $hat{F}_{omega}$ is the free profinite group on countably many generators. Considering $Phi_{n}$, the free metabelian group on $n$ generators, it was also proven that $Cleft(Phi_{2}right)=hat{F}_{omega}$ and $Cleft(Phi_{3}right)supseteqhat{F}_{omega}$. In this paper we prove that $Cleft(Phi_{n}right)$ for $ngeq4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $ngeq3$, in the metabelian case it is between $n=2,3$ and $ngeq4$.