Let $R$ be an integral domain (if it helps i want it to be a polynomial algebra over $\mathbb C$).
We can define the following map: $$ \begin{align} \varphi \colon \mathrm{Gl}_n(R[T, T^{-1}]) & \longrightarrow \mathrm{Gl}_n(R[X, Y, X^{-1}, Y^{-1}]) \\ S(T) & \longmapsto S(X) \cdot S(Y)^{-1} \end{align} $$ and make the following observations:
- If $\varphi(A(T)) = \varphi(B(T))$ then $B(T)^{-1}A(T) \in \mathrm{Gl}_n(R)$, so there exists a matrix $C \in \mathrm{Gl}_n(R)$ with $A(T) = B(T) \cdot C$.\ In particular if $A(X,Y) \in \varphi(\mathrm{Gl}_n(R[T, T^{-1}]))$ then the preimage $\varphi^{-1}(A(X,Y))$ is unique up to multiplication with an element in $\mathrm{Gl}_n(R)$.
- This map is in general not surjective as any element $A(X,Y) \in \varphi(\mathrm{Gl}_n(R[T, T^{-1}]))$ satisfies $\det(A(X,Y) = X^m Y^{-m}$ for some $m \in \mathbb Z$, so any matrix $B(X,Y)\in \mathrm{Gl}_n(R[X, Y, X^{-1}, Y^{-1}])$ with $\det(B(X,Y)) = X^2Y$ cannot lie in the image of $\varphi$.
My question would be
- Is the condition of the determinant sufficient, i.e. given a matrix $B(X,Y) \in \mathrm{Gl}_n(R[X, Y, X^{-1}, Y^{-1}])$ with $\det(B(X,Y))=X^m Y^{-m}$ does there exist a matrix $S(T) \in \mathrm{Gl}_n(R[T, T^{-1}])$ with $\varphi(S(T)) = B(X,Y)$?
- Assume we know that $A(X,Y) = (a_{ij})_{ij} \in \varphi(\mathrm{Gl}_n(R[T, T^{-1}]))$, can one reconstruct a representative $S(T)$ in terms of the $a_{ij}$?