I am slightly confused about the interchangeability of the terms isometry and biholomorphism. This confusion is rooted in the following statement, which I have seen more than once in some discussions on equivalent versions of the Uniformization Theorem.
Suppose $(M,g)$ is a complete, simply connected, $2$-dimensional Riemannian manifold with constant curvature. By the Killing-Hopf theorem, it is isometric to either $\mathbb{R}^2$, $S^2$, or $\mathbb{H}^2$. Identify $M$ as a simply connected Riemann surface and identify $\mathbb{R}^2$, $S^2$, and $\mathbb{H}^2$ as, respectively, $\mathbb{C}$, $\hat{\mathbb{C}}$, and $D_1$. Then $M$ is biholomorphically equivalent to $\mathbb{C}$, $\hat{\mathbb{C}}$, or $D_1$.
To me, an isometry between two Riemannian manifolds $(M,g)$ and $(M',g')$ is a diffeomorphism $f:M\longrightarrow M'$ such that $g=f^*g'$. Thus, it preserves angles and can be described as a conformal mapping.
On the other hand, given two Riemann surfaces $M$ and $N$, a biholomorphism is a bijective map $f:M \longrightarrow N$ such that both $f$ and $f^{-1}$ are holomorphic.
So, in the context of the italicized statement, does it really follow that the isometry turns into a biholomorphism just because we switched from Riemannian manifolds to Riemann surfaces, and because both isometries and biholomorphisms preserve angles? Or is there something more subtle going on here?
from Hot Weekly Questions - Mathematics Stack Exchange
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