Suppose $X$ and $Y$ are smooth affine surfaces over $\mathbb C$. Suppose there is a biholomorphism $f: X\to Y$. Does it follow that $X$ and $Y$ are isomorphic as affine surfaces (i.e. there exists an algebraic isomorphism $g: X\to Y$)?


What if we additionally know that $X$ and $Y$ are rational surfaces?



Biholomorphic but not isomorphic complex affine surfaces?
Let $\overline{C}$ be a complex projective curve of genus $g>0$. Let $C\subset \overline{C}$ be the open affine complement of one (closed) point $p$. The composition $\text{Pic}^0(\overline{C})\to \text{Pic}(\overline{C})\to \text{Pic}(C)$ is an isomorphism. Let $L$ be any nontrivial (geometric) rank $1$ vector bundle over $C$. This is affine, since $C$ is affine and the projection morphism from $L$ to $C$ is affine. This projection morphism equals the image of the Albanese morphism of $L$ (this is a birational invariant for complex projective manifolds, thus extends unambiguously to complex quasi-projective manifolds). The relative tangent bundle of this projection morphism is the pullback from $C$ of a unique invertible sheaf, namely the invertible sheaf associated to $L$. Thus, the relative tangent bundle of the Albanese morphism is not trivial. Hence $L$ is not isomorphic to $\mathbb{A}^1\times C$. Yet the underlying complex manifolds are biholomorphic.


