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 introduce surjective representations of $IA(\Phi_{n})$ onto the group $\ker(GL_{n-1}(\mathbb{Z}[x^{\pm1}])\overset{x\mapsto1}{\longrightarrow}GL_{n-1}(\mathbb{Z}))$ which come via the classical Magnus representation of $IA(\Phi_{n})$. Using this representations combined with some methods and results from Algebraic K-theory, we prove that for every $n\geq4$, $C\left(IA(\Phi_{n}),\Phi_{n}\right)$ contains a product of $n$ copies of the congruence kernel $\ker(\widehat{SL_{n-1}(\mathbb{Z}[x^{\pm1}]})\to SL_{n-1}(\widehat{\mathbb{Z}[x^{\pm1}]}))$ which is central in $\widehat{IA(\Phi_{n})}$. It enables us to show that contrary to free nilpotent cases, $C\left(IA(\Phi_{n}),\Phi_{n}\right)$ is not trivial and not even finitely generated. We note that using some results of this paper we show in an upcoming paper that actually, all the elements of $C\left(IA(\Phi_{n}),\Phi_{n}\right)$ lie in the center of $\widehat{IA(\Phi_{n})}$.